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285lect20 - 1 Lecture 20 Finally we remark that as in the...

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1 Lecture 20 Finally, we remark that, as in the case of a real r , for complex r; D [ e rx ] = re rx : Indeed, let r = a + bi . Then D [ e rx ] = d dx ° e ( a + bi ) x ± = d dx ° e ax e ibx ± = d dx [ e ax (cos bx + i sin bx )] = ae ax (cos bx + i sin bx ) + e ax ( ° b sin bx + bi cos bx ) = ae ax (cos bx + i sin bx ) + bie ax (cos bx + i sin bx ) = ( a + bi ) e ax (cos bx + i sin bx ) = re rx : Fundamental set of solutions By using that D [ e rx ] = re rx , as for real r in cases 1 and 2, we conclude that each complex r 0 = a + bi of multiplicity m 0 contributes to general solution ° C 0 + C 1 x + ::: + C m 0 ° 1 x m 0 ° 1 ± e r 0 x : If r 0 is a complex root of the characteristic equation, then r 0 is also a complex root of the characteristic equation of the same multiplicity (this is because coe¢ cients of the characteristic equation are real). So, r 0 contributes to general solution ° C 0 + C 1 x + ::: + C m 0 ° 1 x m 0 ° 1 ± e r 0 x : Notice that by Euler°s formula, e y 1 = ° C 0 + C 1 x + ::: + C m 0 ° 1 x m 0 ° 1 ± e r 0 x = ° C 0 + C 1 x + ::: + C m 0 ° 1 x m 0 ° 1 ± e ax (cos bx + i sin bx ) : In the mean-time, e y 2 = ° C 0 0 + C 0 1 x + ::: + C 0 m 0 ° 1 x m 0 ° 1 ± e r 0 x =
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